Question: Simplify the following expression and state the conditions under which the simplification is valid. You can assume that $x \neq 0$. $q = \dfrac{x^3 - 4x^2 - 21x}{-2x^3 + 6x^2 + 36x} \div \dfrac{4x - 28}{2x - 14} $
Solution: Dividing by an expression is the same as multiplying by its inverse. $q = \dfrac{x^3 - 4x^2 - 21x}{-2x^3 + 6x^2 + 36x} \times \dfrac{2x - 14}{4x - 28} $ First factor out any common factors. $q = \dfrac{x(x^2 - 4x - 21)}{-2x(x^2 - 3x - 18)} \times \dfrac{2(x - 7)}{4(x - 7)} $ Then factor the quadratic expressions. $q = \dfrac {x(x + 3)(x - 7)} {-2x(x + 3)(x - 6)} \times \dfrac {2(x - 7)} {4(x - 7)} $ Then multiply the two numerators and multiply the two denominators. $q = \dfrac { x(x + 3)(x - 7) \times 2(x - 7)} { -2x(x + 3)(x - 6) \times 4(x - 7)} $ $q = \dfrac {2x(x + 3)(x - 7)(x - 7)} {-8x(x + 3)(x - 6)(x - 7)} $ Notice that $(x + 3)$ and $(x - 7)$ appear in both the numerator and denominator so we can cancel them. $q = \dfrac {2x\cancel{(x + 3)}(x - 7)(x - 7)} {-8x\cancel{(x + 3)}(x - 6)(x - 7)} $ We are dividing by $x + 3$ , so $x + 3 \neq 0$ Therefore, $x \neq -3$ $q = \dfrac {2x\cancel{(x + 3)}\cancel{(x - 7)}(x - 7)} {-8x\cancel{(x + 3)}(x - 6)\cancel{(x - 7)}} $ We are dividing by $x - 7$ , so $x - 7 \neq 0$ Therefore, $x \neq 7$ $q = \dfrac {2x(x - 7)} {-8x(x - 6)} $ $ q = \dfrac{-(x - 7)}{4(x - 6)}; x \neq -3; x \neq 7 $